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Prove that, if:

[tex]y=\frac { { x }^{ n+1 } }{ n+1 } +C[/tex]

Then:

[tex]\\ \\ \frac { dy }{ dx } ={ x }^{ n }[/tex]

By differentiating...

Sagot :

konrad509
[tex]y=\dfrac{x^{n+1}}{n+1}+C\\\\ y'=\left(\dfrac{x^{n+1}}{n+1}+C\right)'\\ y'=\left(\dfrac{x^{n+1}}{n+1}\right)'+C'\\ y'=\dfrac{1}{n+1}\cdot (x^{n+1})'+0\\ y'=\dfrac{1}{n+1}\cdot (n+1)x^n\\ y'=x^n [/tex]
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